MATH 152 Chapter 7.4

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Important Formulas

Force

Force (in Newtons [N]) on an object (in same direction) is the product of mass (in kilograms [Kg]) and acceleration [m/s²]):

$F=m{\frac {\mathrm {d} ^{2}s}{\mathrm {d} t^{2}}}$

(Weight (in pounds [lb]) is a force in itself)

Work

Work is the product of force/weight and distance:

$W=Fd\,\!$

• if $F$ is in Newtons and $d$ is in meters, then $W$ is measured in Joules [J]
• if $F$ is in pounds and $d$ is in feet, then $W$ is measured in foot-pounds [ft-lb] (1 ft-lb ≈ 1.36 J)

Dynamic Work

If force on object at position ($x$) meters (or feet) from origin is a dynamic function ($f(x)$), then work required to move object from $a$ to $b$ on $x$-axis is:

$W=\int _{a}^{b}{f(x)\,\mathrm {d} x}$

...where $f(x)$ is the force, and $\mathrm {d} x$ is the infinitely small change in distance.

Example: Stretching a Spring

A force of 30 N is required to maintain a spring stretched from its natural length of 12 cm to a length of 15 cm. How much work is done in stretching the spring from 12 cm to 20 cm?.

1. Find $k$ in $F=kx$:
   $30=k(.15-.12)$
   $k=1000$
   $F=1000x$
2. Find the force required to hold the spring at a given distance from 12 cm: ($F$ is practically constant)
   $F_{\text{piece}}\approx 1000x_{i}^{*}$
3. Find the work of the piece based on work formula:
   $W_{\text{piece}}=1000x_{i}^{*}\,\Delta x_{i}$
4. Integrate!
   $W=\int _{0}^{.08}{1000x\,\mathrm {d} x}$

Pumping water out of a tank

Water trough is 10 m long and has ends in the shape of inverted isosceles triangles (base = 10, height = 3). Calculate the Work needed to pump all of the water out of the tank if the weight density is $\rho g$ N/L.

Think of the water as a bunch of sheets that are lifted and stacked somewhere else, like in a cartoon.

Slice in the direction of the depth: resulting slice is a rectangular prism:

	$V_{\text{slice}}=lwh\,\!$	$l=10$	(always the length of the trough, regardless of position)
		$w=x$	(more on this later)
		$h=\mathrm {d} y$	(thickness of slice)
∴	$V_{\text{slice}}=10x\,\mathrm {d} y$

Use triangle proportions to get the base of the similar isosceles triangle formed by the water depth:

• $x$ is proportional to the base (8 m) of the triangle
• $y$ is proportional to the height (3 m) of the triangle
• Therefore ${\frac {x}{y}}={\frac {8}{3}};\;x={\frac {8}{3}}y$

$V_{\text{slice}}=10({\frac {8}{3}}y)\,\mathrm {d} y$

Since we are given the weight density ($\rho g$ N/L), we can multiply it by the volume to get the weight/force $F$ of the water:

$F_{\text{slice}}=\rho g(10({\frac {8}{3}}y))\,\mathrm {d} y$

The Formula for work is $W=Fd$. The distance from our arbitrary position $y$ to the top of the tank is $3-y$: when $y=0$, the water has to be lifted 3 m; and when the tank is full (3 m deep), the water doesn't have to be lifted at all. Let's plug this into the formula and clean it up a little:

$W_{\text{slice}}=\rho g(10({\frac {8}{3}}y))(3-y)\,\mathrm {d} y={\frac {80}{3}}\rho g\,(3y-y^{2})\,\mathrm {d} y$

Take the limit of the sum as the norm of the partition approaches zero. In other words, just freakin' integrate from the bottom of the tank to the top of the tank!

$W={\frac {80}{3}}\rho g\int _{0}^{3}{3y-y^{2}\,\mathrm {d} y}$

Suggested Homework

3, 5, 7, 11, 12, 13, 15, 17, 19, 20, 25